The document discusses the chi-square test, a non-parametric statistical test used to compare observed data with expected frequencies. It can be used to test independence and association between two categorical variables. The chi-square test involves calculating a quantity called chi-square from a contingency table of observed and expected frequencies and comparing it to critical values from a chi-square distribution table. If the calculated value exceeds the critical value, the null hypothesis of no relationship can be rejected.

1. M.Prasad Naidu
MSc Medical Biochemistry, Ph.D,.

2.  It is a non parametric test , not based on any
assumption or distribution of any variable.
 It is very useful in research and it is most
commonly used when the data are given in
frequencies .
 It can be used with any data which can be
reduced to proportion or percentages.
 This test involves calculation of quantity
called chi- square , which derived from Greek
letter ( χ)² and pounced as ‘kye’. Cont…

3. It is an alternate test to find the
significance of difference in two or more
than two proportions.
Chi –square test is yet another useful test
which can be applied to find the
significance in the same type of data with
the following advantages .
 cont…

4. To compare the values of two binomial
samples even if they are small such as
incidence of diabetes in 20 obese persons
with 20 non obese persons.
To compare the frequencies of two
multinomial samples such as number of
diabetes and non diabetes in groups
weighing 40-50, 50-60, 60-70, and > 70 kg
s of weight.

5.  Test of association between events in
binomial or multinomial samples .
 Two events often can be tested for their
association such as cancer and smoking .
 Treatment and outcome of disease .
 Vaccination and immunity .
 Nutrition and intelligence .
 Cholesterol and heart disease .
 Weight and diabetes, B P and heart disease.
 To find they are independent of each other or
dependent on each other i.e. associated .

6.  Three essentials to apply chi - square test
are
 A random sample
 Qualitative data
 Lowest frequency not less than 5
 Steps :-
 Assumption of Null Hypothesis (HO)
 Prepare a contingency table and note down
the observed frequencies or data (O)
 cont…

7.  Determine the expected number (E) by
multiplying CT × RT /GT (column total ,row
total and grand total )
 Find the difference between observed and
expected frequencies in each cell (O-E)
 Calculate chi- square value for each cell with
 (O-E)²/E
 Sum up all chi –square values to get the total
chi-square value (χ)² d.f. (degrees of freedom)
 = χ²= ∑(O-E)²/E and d.f. is (c-1) (r-1)

8.  Chi- square test applied in a four fold table
will not give reliable result , with one degree
of freedom.
 If the observed value of any cell is < 5 in
such cases Yates correction can be applied
by subtracting ½ (χ)² = ∑ (O-E-1/2)²/E
 Even with Yates correction the test may be
misleading if any expected value is much
below 5 in such cases Yates correction can
not be applied.
 cont…

9.  In tables larger than 2 × 2 Yates correction can not be
applied .
 The highest value of chi- square χ² obtainable by
chance or worked out and given in (χ)² table at different
degrees of freedom under probabilities (P) such as
0.05, 0.01, 0.001 .
 If calculated value of chi-square of the sample is found
to be higher than the expected value of the table at
critical level of significance i.e. probability of 0.05 the
H O of no difference between two proportions or the H
O of independence of two characters is rejected.
 If the calculated value is lower the hypothesis of no
difference is accepted.

10.  Groups Died Survived Total
 A (a) 10 (b) 25 35
 B (c) 5 (d) 60 65
 Total 15 85 100
 Expected value can be computed with CT×RT/GT
for (a) 15×35/100= 5.25
 (b) 85×35/100= 29.75
 (c) 15×65/100= 9.75
 (d) 85×65/100= 55.25

11.  χ²= ∑(O-E)²/E d.f. =( c-1) (r-1)

 (A) =(10-5.25)²/5.25 = (4.75)²/5 = 4.30
 (B) = (25-29.75)²/29.75 =(5)²/29.75 = 0 .76
 (C) = (5-9.75)²/9.75 = (5)²/9.75 = 2.31
 (D) = (60-55.25)²/55.25 = (5)²/55.25 = 0.40
7.77
The calculated chi- square value is 7.77 is more than
Chi- square table value with 1 d.f. At 0.01, 6.64
which significant. It can be said that there is
significant difference between two groups.

12.  Chi- square can be calculated in the following
way also.
 (χ)² = ( ad-bc)²×N
(a b × c d ×ac ×b d) = (600-125 )²
( 35× 65× 15×
85)
= (475)² x100 225625 00 /2900625
(2900625) = 7.77
Here the calculated chi-square value7.77 is >
6.64 at
0.01 with 1 d.f. shows significant difference b/n